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Δεύτερο Θεώρημα Noether
Definition Noether's theorem E. Noether, "Invariante Variationsprobleme", Nachr. d. König. Gesellsch. d. Wiss. zu Göttingen, 235-257 (1918); arXiv:physics/0503066 (English translation) states that for every continuous symmetry in classical mechanics there exists a conservation law. Introductory formulation Consider the effect of a general transformation of both the coordinates and the fields: Generally, the new coordinate x'\, is a function of the old coordinate x\, , and both coordinates label the same point P\, . We restrict ourselves to cases where the new field at x'\, is a function of the old field at x\, : : \Phi'(x') = \mathcal{F}(\Phi(x))\, . First, we restrict ourselves to some compact submanifold B\, of M\, . Suppose, after performing the transformation, and without invoking the equations of motion, we find that : S_B' - S_B = \int_B\!d^dx\, \partial_\mu f^\mu\, . Because the inclusion of a surface term does not affect the derivation of the equations of motion, the transformation is thus a symmetry of the action. Now, i.e., we do not shift our integration variables to be compatible with the fields. Instead, we will make this shift implicit with the field deformations, defining \Delta\Phi(x) = \Phi'(x) - \Phi(x)\, . We may obtain an expression for S_B'-S_B\, a different way, by using the equations of motion, to find Now \Delta\left[ \partial_\mu\Phi(x)\right] = \left.\frac{\partial}{\partial(x^{\prime\mu})} \Phi'(x')\right|_{x' = x} - \partial_\mu \Phi(x) = \partial_\mu \Phi'(x)- \partial_\mu \Phi(x) = \partial_\mu\Delta\Phi(x)\, , hence the commutator vanishes. We now have two expressions for S_B'-S_B\, , This is true for any choice of B\, , hence : \partial_\mu\left[ f^\mu - \left( \frac{\partial\mathcal{L}}{\partial(\partial_\mu\Phi)} \Delta\Phi\right) \right] = 0\, . Thus the Noether current : J^\mu \equiv f^\mu - \left( \frac{\partial\mathcal{L}}{\partial(\partial_\mu\Phi)} \Delta\Phi\right)\, is conserved, provided that the equations of motion hold. Detailed formulation Under the transformation the action becomes Now the infinitesimal form of the transformation can be written in terms of parameters \omega_a\, as Using \det(1+\delta M) = \exp\operatorname{tr} \ln(1+\delta M) \approx \exp \operatorname{tr} \delta M \approx 1 + \operatorname{tr} \delta M\, , the Jacobian becomes : \left| \frac{\partial\mathbf{x}'}{\partial\mathbf{x}}\right| \approx 1 + \partial_\mu \left( \omega_a \frac{\delta {x^\mu}'}{\delta \omega_a} \right)\, , while : \frac{\partial x^\nu}{\partial {x^\mu}'} = \delta^\nu_\mu - \partial_\mu\left( \omega_a \frac{\delta {x^\nu}'}{\delta \omega_a} \right)\, . Το Δεύτερο Θεώρημα Generally, : \delta\mathcal{L} = \delta x^\mu \partial_\mu \mathcal{L} + \delta_0 \mathcal{L}\, , where \delta_0\mathcal{L}(x) = \mathcal{L}'(x) - \mathcal{L}(x)\, is the functional change in \mathcal{L}\, , i.e., the change keeping the coordinates (not the points) fixed. Note that \delta_0\, commutes with \partial_\mu\, . Then, : \delta\mathcal{L} = \delta x^\mu \partial_\mu \mathcal{L} + \frac{\partial \mathcal{L}}{\partial\Phi} \delta _0\Phi + \frac{\partial \mathcal{L}}{\partial(\partial_\mu\Phi)} \delta_0 \partial_\mu \Phi\, Thus For classical solutions, the second term vanishes by the equations of motion. Now, \delta_0 \Phi = \delta\Phi - \delta x^\mu \partial_\mu \Phi\, , so Evidently, if the transformation is a symmetry of the action, then S'-S=0\, for any infinitesimal parameters \omega^a\, , so that we have the following continuity (conservation) equation : \partial_\mu J^\mu_a = 0\, , where : J^\mu_a = \left( \delta^\mu_\nu \mathcal{L} - \frac{\partial \mathcal{L}}{\partial(\partial_\mu\Phi)} \partial_\nu \Phi \right) \frac{\delta x^\nu}{\delta \omega^a} + \frac{\partial \mathcal{L}}{\partial(\partial_\mu\Phi)} \frac{\delta\mathcal{F}}{\delta\omega^a}\, is a conserved current known as the Noether current associated with the symmetry. Efficient computation One way of obtaining the Noether current that is often most efficient (although not particularly transparent) is to recognize that if some symmetry transformation \Phi \to \Phi + \delta\Phi\, leaves the action invariant, then : S\Phi = S+ \varepsilon \delta\Phi\, , where \varepsilon\, is some infinitesimal constant. Now, if we let \varepsilon\, be spacetime dependent, then to linear order in \varepsilon(x)\, , the variation in the action should depend on derivatives of \varepsilon(x)\, . We can then, using integration by parts to obtain the form : \delta S = \int\!d^dx\, J^\mu(x) \partial_\mu \varepsilon(x)\, , for some J^\mu(x)\, . By further integration by parts, we obtain : \delta S = -\int\!d^dx\, \left( \partial_\mu J^\mu(x)\right) \varepsilon(x)\, . However, if \Phi\, is a classical solution, then any variation around \Phi\, should leave the action invariant, so : \delta S = -\int\!d^dx\, \left( \partial_\mu J^\mu(x)\right) \varepsilon(x) = 0,\, \forall \varepsilon(x)\, . This can only be true if : \partial_\mu J^\mu = 0\, Thus by using a test function \varepsilon(x)\, , and performing the variation and necessary integration by parts, we can always read off the current from the coefficient of \partial_\mu \varepsilon(x)\, in : \delta S = \int\!d^dx\, J^\mu(x) \partial_\mu \varepsilon(x)\, . for some J^\mu(x)\, . References Κατηγορία:Κλασσική Μηχανική